# Kinematic problems

1. Sep 27, 2005

The human body can survive a negative acceleration trauma incident if the magnitude of the acceleration is less than 250 m/s^2. If you are in an automobile accident at an initial speed of 96 km/h and are stopped by an airbag that inflates from the dashboard, over what distance must the airbag stop you for you to survive the crash?

So I know that $v_{0} = 96$, $v_{x} = 0$ and $a_{x} = 250$. So is it correct to say $v_{x} = v_{x}_{0} + a_{x}t$ to find the time, or $0 = 96-250t$ and $t = 0.384 sec$? Then you use $x-x_{0} = v_{x}_{0}t + \frac{1}{2}a_{x}t^{2}$ and you get the distance to be $18.432 m$

Is this correct?

Thanks

2. Sep 27, 2005

any ideas?

3. Sep 27, 2005

### Leong

change 96 km/h to m/s.

4. Sep 27, 2005

### Jameson

I would do the above suggestion and use this equation... it's faster.

$$V_{f}^2=V_{0}^2+2ad$$

5. Sep 27, 2005